Stable and unstable nuclei

AQA·Physics·3.2.1.2

Strong nuclear force

The nucleus contains protons and neutrons, collectively called nucleons. Protons repel one another because they have the same positive electric charge. The strong nuclear force acts between nucleons and helps hold the nucleus together against this electrostatic repulsion.

The strong nuclear force depends on the separation between nucleons. It is attractive at separations from about $$0.5\ \text{fm}$$ up to about $$3\ \text{fm}$$, pulling nearby nucleons towards one another. At very short separations, below about $$0.5\ \text{fm}$$, it becomes repulsive and pushes them apart. A femtometre ($$\text{fm}$$) is $$10^{-15}\ \text{m}$$.

The force has a short range: its attractive effect is significant only between nucleons that are close together. Beyond approximately $$3\ \text{fm}$$, it is no longer significant.

Alpha decay

In alpha decay, an unstable nucleus emits an alpha particle: a cluster of two protons and two neutrons. The original nucleus is called the parent nucleus, and the nucleus left behind is called the daughter nucleus.

The mass number is the total number of protons and neutrons in a nucleus. Because the alpha particle carries away four of these particles, the daughter’s mass number is four lower than the parent’s. Its atomic number, the number of protons, is two lower. The daughter is therefore a different element from the parent.

Alpha decay equation

If the parent nucleus has mass number $$A$$ and atomic number $$Z$$, its alpha decay can be written as:

ZAX→Z−2A−4Y+24He{}^{A}_{Z}X \rightarrow {}^{A-4}_{Z-2}Y + {}^{4}_{2}\mathrm{He}

Here, $$X$$ represents the parent nucleus, $$Y$$ the daughter nucleus, and $${}^{4}_{2}\mathrm{He}$$ the alpha particle. An alpha particle is a helium nucleus.

Both mass number and atomic number are conserved across the equation: the mass numbers on the right add to $$(A-4)+4=A$$, and the atomic numbers add to $$(Z-2)+2=Z$$.

Beta-minus decay

In beta-minus decay, a neutron in an unstable nucleus changes into a proton. An electron and an electron antineutrino are emitted. The emitted electron is the beta-minus particle; it is produced in the decay rather than coming from an electron shell.

The nucleus gains one proton and loses one neutron. Its total number of protons and neutrons, the mass number, therefore stays the same. Its number of protons, the atomic number, increases by one, so the daughter nucleus is a different element.

ZAX→Z+1AY+−10e+νˉe{}^{A}_{Z}X \rightarrow {}^{A}_{Z+1}Y + {}^{0}_{-1}e + \bar{\nu}_e

Here, $$X$$ is the parent nucleus and $$Y$$ is the daughter nucleus. The symbols $$A$$ and $$Z$$ represent the parent's mass number and atomic number. The electron is written as $${}^{0}_{-1}e$$, and $$\bar{\nu}_e$$ represents the electron antineutrino.

Energy in beta-minus decay

The emitted electron does not always carry the same energy. The available energy is shared between the electron, the antineutrino and the recoiling daughter nucleus, and the share received by the electron can vary from one decay to another.

The neutrino was hypothesised to account for conservation of energy in beta decay: energy not carried by the electron or the recoiling nucleus is carried away by the antineutrino.